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Why A() is a scalarField ?

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Old   October 16, 2023, 04:50
Default Why A() is a scalarField ?
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Hello all,


looking at the fvMatrix class (but you also see it in all the incompressible solvers), I noticed that the A() method return a volScalarField, while the H() a volumetric field based on the type.
I would have expected a field base on type also for the A(), or the matrix diagonal for every component.
Since I am wrong, could someone explain it to me ?
Thank you
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Old   October 18, 2023, 06:02
Default Just a convenient note
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In my opinion, A() comes from
MU=-\nabla p
and
MU=AU-H
where A is a diagonal matrix, H is a non-diagonal matrix
For the sake of simplicity, we can treat A as a fvScalarField, namely ignoring zeros in A matrix.
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Old   October 18, 2023, 10:59
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Quote:
Originally Posted by ZX-zha View Post
In my opinion, A() comes from
MU=-\nabla p
and
MU=AU-H
where A is a diagonal matrix, H is a non-diagonal matrix
For the sake of simplicity, we can treat A as a fvScalarField, namely ignoring zeros in A matrix.
Just to add to this response, look at the fvMatrix code and you will see that A() just returns the matrix diagonal coefficients (dimensionless) divided by the cell volume. Hence, this is always a volScalar. By contrast, the dimensions of H() are the same as the dimensions of the matrix source term, and so depend on the matrix equation that is being solved - the post above shows one example where it has the dimensions of a volVectorField, but it could be anything.

Last edited by Tobermory; October 18, 2023 at 12:01.
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Old   October 18, 2023, 12:07
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Quote:
Originally Posted by ZX-zha View Post
In my opinion, A() comes from
MU=-\nabla p
and
MU=AU-H
where A is a diagonal matrix, H is a non-diagonal matrix
For the sake of simplicity, we can treat A as a fvScalarField, namely ignoring zeros in A matrix.
I think a better description of the matrix equation, for OpenFOAM, is as follows:
M U^*= Q^n -\nabla p^n

where Q is the source term. Now define H as the sum of the source term minus the off-diagonal terms of MU, then
AU^*=H -\nabla p

where AU contains the remaining diagonal terms from MU. Using this terminology, you should be able to follow through the coding in the pEqn more easily.
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